Groups, rings and fields

A group is a set $G$ with a binary operation, written $+$ here, satisfying four properties:

1. Closure. For all $a, b \in G$, $a + b \in G$.
2. Associativity. For all $a, b, c \in G$, $(a + b) + c = a + (b + c)$.
3. Identity element. There exists $0 \in G$ such that $0 + a = a + 0 = a$ for every $a$.
4. Inverse elements. For every $a \in G$, there exists $-a \in G$ with $a + (-a) = (-a) + a = 0$.

The operation does not have to be literal addition. It can be multiplication, or any binary operation meeting these four requirements. When the operation is also commutative ($a + b = b + a$ for all $a, b$), the group is an abelian group, also called a commutative group.

A ring adds a second operation, multiplication, on top of the additive group structure:

1. Closure. $a + b$ and $a \cdot b$ are both in the set, for all $a, b$.
2. Associativity. $(a+b)+c = a+(b+c)$ and $(a \cdot b) \cdot c = a \cdot (b \cdot c)$.
3. Additive identity. $a + 0 = a$ for all $a$.
4. Additive inverse. For every $a$, there is $-a$ with $a + (-a) = 0$.
5. Commutativity of addition. $a + b = b + a$.
6. Distributive laws. $a \cdot (b + c) = a \cdot b + a \cdot c$ and $(a+b) \cdot c = a \cdot c + b \cdot c$.

Two things a ring does not require: multiplication does not have to be commutative ($a \cdot b = b \cdot a$ can fail), and there does not have to be a multiplicative identity. $\mathbb{Z}$ satisfies all six properties, so $\mathbb{Z}$ is a ring.

A commutative ring is a ring with one more property:

7. Commutativity of multiplication. $a \cdot b = b \cdot a$ for all $a, b$.

$\mathbb{Z}$ satisfies this too, so $\mathbb{Z}$ is a commutative ring. A ring with a multiplicative identity (an element $1$ with $a \cdot 1 = a$) is called a ring with unity. An integral domain is a commutative ring with unity that additionally has no zero divisors: $a \cdot b = 0$ forces $a = 0$ or $b = 0$. $\mathbb{Z}$ is an integral domain. The supplement introduces the term as part of the algebraic-structure hierarchy but does not work through a separate proof for it, so no additional worked example is given here beyond noting where it sits: strictly between commutative ring and field.

A field adds the strongest requirement of all: a multiplicative inverse for every nonzero element.

8. Multiplicative inverse. For every $a \neq 0$, there exists $a^{-1}$ with $a \cdot a^{-1} = 1$.

A field is a commutative ring with unity where that eighth property also holds, supporting addition, subtraction, multiplication, and division by anything except zero.

A prime field, written $F_p$ or $\mathrm{GF}(p)$, is the set $\{0, 1, \ldots, p-1\}$ with addition and multiplication taken modulo a prime $p$. The lecture works through $F_7 = \mathrm{GF}(7) = \{0,1,2,3,4,5,6\}$, checking every field axiom directly against numeric examples.

A prime field is only a field because $p$ is prime. The general set $\mathbb{Z}_n = \{0, 1, \ldots, n-1\}$ under addition and multiplication modulo $n$ is always a commutative ring, but it is only a field when $n$ is prime. $\mathbb{Z}_8$ makes the gap concrete: $8$ is not prime, and $2$ turns out to have no multiplicative inverse mod $8$.